Problem: Solve for $q$, $ \dfrac{7}{8q + 8} = \dfrac{3q + 7}{6q + 6} + \dfrac{10}{2q + 2} $
Solution: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8q + 8$ $6q + 6$ and $2q + 2$ The common denominator is $24q + 24$ To get $24q + 24$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{7}{8q + 8} \times \dfrac{3}{3} = \dfrac{21}{24q + 24} $ To get $24q + 24$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{3q + 7}{6q + 6} \times \dfrac{4}{4} = \dfrac{12q + 28}{24q + 24} $ To get $24q + 24$ in the denominator of the third term, multiply it by $\frac{12}{12}$ $ \dfrac{10}{2q + 2} \times \dfrac{12}{12} = \dfrac{120}{24q + 24} $ This give us: $ \dfrac{21}{24q + 24} = \dfrac{12q + 28}{24q + 24} + \dfrac{120}{24q + 24} $ If we multiply both sides of the equation by $24q + 24$ , we get: $ 21 = 12q + 28 + 120$ $ 21 = 12q + 148$ $ -127 = 12q $ $ q = -\dfrac{127}{12}$